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It looks like you've uploaded an image containing a problem about estimating population using the Capture-Recapture method, commonly known as the Lincoln-Petersen estimator. Since I can't view images directly, I’ll solve the problem based on the text you've provided.

Let’s go through the problem step by step and solve it.

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Problem Summary:

A sample of 500 people from a town is captured (marked), and then a second sample of 400 people is taken later. In this second sample, 100 people are found to have been marked in the first sample.

We are to:

1. Estimate the total population size using the Lincoln-Petersen index.
2. Determine the number of people who must be sampled in the second round so that the standard deviation of the estimate is less than or equal to 100.
3. Calculate the 95% confidence interval for the population estimate.

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Step 1: Estimate Population Size (Lincoln-Petersen Estimator)

The formula for the Lincoln-Petersen estimator is:

\[
\hat{N} = \frac{M \cdot C}{R}
\]

Where:
- \( M \) = number of individuals marked in the first sample = 500
- \( C \) = number of individuals captured in the second sample = 400
- \( R \) = number of recaptured (marked) individuals in the second sample = 100

So,

\[
\hat{N} = \frac{500 \cdot 400}{100} = \frac{200,000}{100} = 2000
\]

✔ Estimated population size: 2000

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Step 2: Find Sample Size for Second Capture So That Standard Deviation ≤ 100

The variance of the Lincoln-Petersen estimator is approximately:

\[
\text{Var}(\hat{N}) = \frac{M C (M - R)(C - R)}{R^2 (R + 1)}
\]

But a more practical approximation (especially when \( R \) is large) uses:

\[
\text{Var}(\hat{N}) \approx \frac{M C (M - R)(C - R)}{R^2 (R + 1)}
\]

However, since we want to control the standard deviation of the estimate, we need to find how large \( C \) (second sample size) should be such that:

\[
\text{SD}(\hat{N}) \leq 100
\]

Let’s use the approximate variance formula:

\[
\text{Var}(\hat{N}) \approx \frac{M C (M - R)(C - R)}{R^2 (R + 1)}
\]

But note: \( R \) depends on \( C \). We assume that the proportion of marked individuals in the second sample remains roughly constant. Since in the original case, \( R = 100 \) out of \( C = 400 \), the recapture rate is:

\[
p = \frac{R}{C} = \frac{100}{400} = 0.25
\]

So, if we let \( C \) be the new second sample size, then expected \( R = 0.25C \)

Now, plug into the variance formula:

\[
\text{Var}(\hat{N}) = \frac{M C (M - R)(C - R)}{R^2 (R + 1)}
\]

Substitute:
- \( M = 500 \)
- \( R = 0.25C \)

Then:

\[
\text{Var}(\hat{N}) = \frac{500 \cdot C \cdot (500 - 0.25C) \cdot (C - 0.25C)}{(0.25C)^2 (0.25C + 1)}
= \frac{500 C (500 - 0.25C)(0.75C)}{(0.0625 C^2)(0.25C + 1)}
\]

Simplify numerator:

\[
500 \cdot C \cdot (500 - 0.25C) \cdot 0.75C = 500 \cdot 0.75 \cdot C^2 (500 - 0.25C) = 375 C^2 (500 - 0.25C)
\]

Denominator:

\[
0.0625 C^2 (0.25C + 1)
\]

So:

\[
\text{Var}(\hat{N}) = \frac{375 C^2 (500 - 0.25C)}{0.0625 C^2 (0.25C + 1)} = \frac{375 (500 - 0.25C)}{0.0625 (0.25C + 1)}
\]

Simplify constants:

\[
\frac{375}{0.0625} = 6000
\]

So:

\[
\text{Var}(\hat{N}) = \frac{6000 (500 - 0.25C)}{0.25C + 1}
\]

We want:

\[
\sqrt{\text{Var}(\hat{N})} \leq 100 \Rightarrow \text{Var}(\hat{N}) \leq 10,000
\]

So:

\[
\frac{6000 (500 - 0.25C)}{0.25C + 1} \leq 10,000
\]

Multiply both sides:

\[
6000(500 - 0.25C) \leq 10,000(0.25C + 1)
\]

Compute:

Left: \( 6000 \cdot 500 = 3,000,000 \); \( 6000 \cdot 0.25C = 1500C \)

So:

\[
3,000,000 - 1500C \leq 2500C + 10,000
\]

Bring all terms to one side:

\[
3,000,000 - 10,000 \leq 2500C + 1500C
\Rightarrow 2,990,000 \leq 4000C
\]

\[
C \geq \frac{2,990,000}{4000} = 747.5
\]

So, \( C \geq 748 \)

✔ Answer: At least 748 individuals must be sampled in the second round.

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Step 3: Compute 95% Confidence Interval

For the Lincoln-Petersen estimator, a common confidence interval is given by the Chao-Lee or bootstrap methods, but a simple approximation uses:

\[
\hat{N} \pm z \cdot \sqrt{\text{Var}(\hat{N})}
\]

Using \( z = 1.96 \) for 95% CI.

From earlier, with \( M = 500 \), \( C = 400 \), \( R = 100 \):

Use the approximate variance formula:

\[
\text{Var}(\hat{N}) = \frac{M C (M - R)(C - R)}{R^2 (R + 1)}
= \frac{500 \cdot 400 \cdot (500 - 100) \cdot (400 - 100)}{100^2 \cdot 101}
\]

Calculate step-by-step:

- \( 500 \cdot 400 = 200,000 \)
- \( 500 - 100 = 400 \)
- \( 400 - 100 = 300 \)
- Numerator: \( 200,000 \cdot 400 \cdot 300 = 200,000 \cdot 120,000 = 24,000,000,000 \)
- Denominator: \( 100^2 \cdot 101 = 10,000 \cdot 101 = 1,010,000 \)

So:

\[
\text{Var}(\hat{N}) = \frac{24,000,000,000}{1,010,000} \approx 23,762.37
\]

Then:

\[
\text{SD}(\hat{N}) = \sqrt{23,762.37} \approx 154.15
\]

Now, 95% CI:

\[
\hat{N} \pm 1.96 \cdot 154.15 = 2000 \pm 302.14
\]

So:

\[
\text{CI} = [1697.86, 2302.14]
\]

Rounded to nearest whole number:

✔ 95% Confidence Interval: [1698, 2302]

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✔ Final Answers:

1. Estimated population: \( \boxed{2000} \)

2. Minimum second sample size for SD ≤ 100: \( \boxed{748} \)

3. 95% Confidence Interval: \( \boxed{[1698,\ 2302]} \)

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